- On 9 Mar 2004 at 14:07:14, "Daniel Rossi de Campos" (Daniel.Campos.at.anvisa.gov.br) sent the message
Dear All,

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Sorry, but a need to make this basic query:

Could anyone say how to calculate the concentrations from a polinomial

regression (Y = ax2 + bx + c where: y = area ratio and x =

concentration (ng/mL) in a bioanalytical assay?

Thanks,

Daniel

P.S n I have tried to use Baskara theorema, but I could not get

adequate results. - On 9 Mar 2004 at 16:52:11, Richard Sams (sams.at.mail.atl.ohio-state.edu) sent the message

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Daniel,

You can perform this calculation easily in an Excel spreadsheet after

you have obtained the polynomial expression that best describes the

relationship between the area ratio and the concentrations of your

calibrators by using the "Goal Seek" function in the drop down menu

under "Tools". You will be asked to input the following:

"Set cell:" You need to enter the designation of the cell in which the

polynomial equation is expressed

"To value:" You need to enter the numerical value of the area ratio

"By changing cell:" You need to enter the designation of the cell in

which you want the concentration to be reported.

After you have provided this information, click on "OK" in the dialog

box and the concentration will appear in the cell that you have

designated for the answer.

Sincerely,

Richard Sams - On 10 Mar 2004 at 08:45:59, Michael.Gassen.at.4sc.com sent the message

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Dear Daniel,

For a square polynomal function, you have a simple option in Excel:

1. Make a graph of your data (x-y type)

2. Add a trendline, use the option "put equation in graph"

For anything more complicated, you can use the solver add-in (needs

installation via add-in manager), which will do numerical regression for

basicly every equation. You just make a spreadsheet with your measured

data

Y. For e.g. an equation like Y = ax3 + bx2 + cx + d, you create fou

cells

where you type your guestimates for the parameters a-d. Based on these

parameters and your equation, you calculate the "theoretical" values of

Y(th) and the sum of square differences sum((Y-Y(th))^2). Then you

minimize

this sum with the solver with variation of the parameters (a-d).

good luck,

Michael

Michael Gassen

Head of Pharmacology and Preclinical Development

4SC AG Fon: +49 89 700763-0

Am Klopferspitz 19a Fax: +49 89 700763-29

82152 Martinsried E-Mail: gassen.-a-.4sc.com

Germany Internet: www.4sc.com - On 10 Mar 2004 at 10:50:37, Pascal.Delrat.-a-.uk.netgrs.com sent the message

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Hi,

since C and Y are constant in your eqution,

why not try to resolve ax2 + bx + c -Y =0

You should end up on ax2 + bx + K =0 with K =c-Y since Y should be > to

c

you should end up on :ax2 + bx - K =0

Then resolve the equation for x with finding the two roots with one

might be

negative (that you reject) and the other one your concentration.

It would help to have the constants of your equation to know which

solution

is adequate.

Do not if this helps,

Regards,

Pascal - On 10 Mar 2004 at 10:55:10, David_Critchley.aaa.eisai.net sent the message

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Dear Daniel,

There will be 2 outcomes possible for such a quadratic function (so

using

'goal seek' may not be the best way to solve) - obviously, only one will

correspond to your concentration (you will know which is correct). To

work

out the two possible values for x, use the following formulae (where Y

= ax

2 + bx + c):

x1 = - (b/(2a)) - {square root [ ((Y-c)/a) + (b/(2a))2]}

x2 = - (b/(2a)) + {square root [ ((Y-c)/a) + (b/(2a))2]}

I hope this is helpful.

Cheers,

Dave - On 15 Mar 2004 at 10:17:05, "Henri Merdjan" (hmerdjan.-a-.wanadoo.fr) sent the message
Dear members,

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I am writing regarding a recent debateabout the use of polynomials for

calibration purpose. Let me add a few comments on it.

I haven't retained the original messages. However, if my recollection

is correct, several members provided the right answers. More

specifically, two solutions exist for ax2 + bx +c =0. By the way,

somebody wrote that one is positive and the other is negative. This is

not always right, both may be positive, or both negative.

The existence of 2 solutions illustrate an undesirable property of

polynomials in calibrations: they are non-monotonic functions. For this

reason, I would like to suggest 2 alternatives. Bothare based

onmonotonic functions.

1. One optionwould beto use linear regression on log-transformation

data (on both axes). LnY=a*LnX + b. This option, however,suffers 2

main (hidden) drawbacks. One is an underlying assumption of a

constant-CV error model, rarely applicable to bioanalytical data.

Another one is that the model is essentially forced through the origin.

This is obvious after antilog transformation: Y = exp(b)*X**a.The

main value of this approachis its straightforward implementation.

2. The second option would be tu use NONlinear regression on

UNtransformed data,usingthe monotonic model: Y = a*X**n + b. By

contrast with option 1, these is no specific assumption about the error

model (feel free to use the most appropriate weighting function, if

any) and the model is not forced through the origin thanks to the

intercept "b". On the other hand, implementation is not straightforward

asit requires a nonlinear regression module (e.g. Excel solver).

Enjoy the testing!

Henri

Henri Merdjan

Consultant to the Life Science Industries

PK and PK/PD specialist

34, rue des Bergers

75015 Paris

France

hmerdjan.-at-.wanadoo.fr - On 18 Mar 2004 at 09:55:43, "J.H.Proost" (J.H.Proost.at.farm.rug.nl) sent the message

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Dear Henri,

Thank your for your suggestions for alternatives for the polynomial

regression.

Your comments with respect to the properties of linear regression

applied to log-transformed data are correct, but I do not fully agree

with your conclusions:

- the underlying assumption of a constant-CV error may be not fully

justified, but this also applies to the assumption of a constant error

underlying the use of linear regression on untransformed and unweighted

data. In real data, the situation is usually somewhere between.

- forcing the model through the origin is not a bad idea. In case one

knows that there a blank sample produces a response, one can make

corrections for it. Anyhow, if this is the case, also linear regression

may give rise to inaccuracies, and one has to be quite sure that the

calibration curve is linear. Please note that after log-transformation

the concentration zero is infinitely far away from the lowest

calibration sample. Nobody would ever make such an extrapolation! So it

is not really important that the calibration curve in the untransformed

scale is forced through the origin.

In addition, linear regression requires that the X-values are regularly

distributed over the range. This is usually not the case. After

logarithmic transformation this requirement may be fulfilled better (in

practice, again, the situation is probably somewhere between).

With respect to the use NONlinear regression and the monotonic model: Y

a*X**n + b: If data are not weighted, there is certainly a specified

underlying (hidden) assumption about the error model. I would not

recommend to use this method without careful analysis of the results,

e.g. in residuals plots, quite similar to common practice in

pharmacokinetics.

Best regards,

Hans

Johannes H. Proost

Dept. of Pharmacokinetics and Drug Delivery

University Centre for Pharmacy

Antonius Deusinglaan 1

9713 AV Groningen, The Netherlands

tel. 31-50 363 3292

fax 31-50 363 3247

Email: j.h.proost.-a-.farm.rug.nl

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